146. LRU Cache

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4
什么是LRU?
LRU(least recently used)最近最少使用。
假设 序列为 4 3 4 2 3 1 4 2
物理块有3个 则
首轮 4调入内存 4
次轮 3调入内存 3 4
之后 4调入内存 4 3
之后 2调入内存 2 4 3
之后 3调入内存 3 2 4
之后 1调入内存 1 3 2(因为最少使用的是4,所以丢弃4)
之后 4调入内存 4 1 3(原理同上)
最后 2调入内存 2 4 1
import java.util.HashMap;


public class LRUCache {
    private HashMap<Integer, DoubleLinkedListNode> map
            = new HashMap<Integer, DoubleLinkedListNode>();
    private DoubleLinkedListNode head;
    private DoubleLinkedListNode end;
    private int capacity;
    private int len;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        len = 0;
    }

    public int get(int key) {
        if (map.containsKey(key)) {
            DoubleLinkedListNode latest = map.get(key);
            removeNode(latest);
            setHead(latest);
            return latest.val;
        } else {
            return -1;
        }
    }

    public void removeNode(DoubleLinkedListNode node) {
        DoubleLinkedListNode cur = node;
        DoubleLinkedListNode pre = cur.pre;
        DoubleLinkedListNode post = cur.next;

        if (pre != null) {
            pre.next = post;
        } else {
            head = post;
        }

        if (post != null) {
            post.pre = pre;
        } else {
            end = pre;
        }
    }

    public void setHead(DoubleLinkedListNode node) {
        node.next = head;
        node.pre = null;
        if (head != null) {
            head.pre = node;
        }

        head = node;
        if (end == null) {
            end = node;
        }
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {
            DoubleLinkedListNode oldNode = map.get(key);
            oldNode.val = value;
            removeNode(oldNode);
            setHead(oldNode);
        } else {
            DoubleLinkedListNode newNode =
                    new DoubleLinkedListNode(key, value);
            if (len < capacity) {
                setHead(newNode);
                map.put(key, newNode);
                len++;
            } else {
                map.remove(end.key);
                end = end.pre;
                if (end != null) {
                    end.next = null;
                }

                setHead(newNode);
                map.put(key, newNode);
            }
        }
    }
}

class DoubleLinkedListNode {
    public int val;
    public int key;
    public DoubleLinkedListNode pre;
    public DoubleLinkedListNode next;

    public DoubleLinkedListNode(int key, int value) {
        val = value;
        this.key = key;
    }

}

  

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原文地址:https://www.cnblogs.com/clarencezzh/p/10919035.html